I'd like to ask about bringing the derivative inside the expected value.
The context is proving that if $\mathbb{E}X < \infty$, then $\mathbb{E}X = \frac{1}{i}\frac{\mathrm{d}\phi_X(t)}{\mathrm{d}t}|_{t = 0}$. It is the first proposition here where $n = 1$. The argument begins with
$$ \begin{align} \frac{\mathrm{d}\phi_X(t)}{\mathrm{d}t} &= \frac{\mathrm{d}\mathbb{E}e^{itX}}{\mathrm{d}t} \\ &= \mathbb{E}\frac{\mathrm{d}e^{itX}}{\mathrm{d}t}. \end{align} $$
I see various references mention Lebesgue's dominated convergence theorem. I'm trying to use the notation from slide 3 here:
$$ \lim_{y\rightarrow y_0}\int h(x, y)\mathrm{d}x = \int \lim_{y\rightarrow y_0}h(x, y)\mathrm{d}x, $$
where $h$ satisfies appropriate conditions.
What is $h$ when bringing the derivative inside the expected value? My attempt for absolutely continuous $X$:
$$ \begin{align} h(x, t) &= \frac{e^{i(t + y)x} - e^{itx}}{y}. \\ \frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}e^{itX} &= \frac{\mathrm{d}}{\mathrm{d}t}\int e^{itx}f_X(x) \mathrm{d}x \\ &= \lim_{y\rightarrow 0} \frac{\int e^{i(t + y)x}f_X(x) \mathrm{d}x - \int e^{itx}{f_X}(x) \mathrm{d}x}{y} \\ &= \lim_{y\rightarrow 0}\int \frac{e^{i(t + y)x} - e^{itx}}{y}f_X(x) \mathrm{d}x \\ &= \lim_{y\rightarrow 0}\int h(x, t)f_X(x) \mathrm{d}x \\ &= \int \lim_{y\rightarrow 0}\frac{e^{i(t + y)x} - e^{itx}}{y}f_X(x) \mathrm{d}x \\ &= \int \frac{\mathrm{d}}{\mathrm{d}t}e^{itx}f_X(x) \mathrm{d}x \\ &= \mathbb{E}\frac{\mathrm{d}}{\mathrm{d}t} e^{itX}, \end{align} $$
but then how do you bound $|h|$?
Including the density in $h$ allows the application of $\mathbb{E}|X| < \infty$.